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JEE Advanced 2017 Paper 1 · Mathematics · Differentiation and Application of Derivatives · medium

JEE Advanced 2017 Paper 1 Mathematics: Paragraph 2

Column 1 gives four functions through their derivatives, on the stated domains (with an initial value), Column 2 gives the abscissa of the unique global extremum, and Column 3 gives the corresponding extreme value. Each function has exactly one entry in each of the other two columns, which are all used exactly once.

| Column 1 (function) | Column 2 (point of global extremum) | Column 3 (extreme value) |
|---|---|---|
| (I) f′(x)=1+ln⁡x, x>0, f(1)=0f'(x)=1+\ln x,\ x>0,\ f(1)=0 | (i) x=1x=1 | (P) 4e−24e^{-2} |
| (II) f′(x)=1−ln⁡xx2, x>0, f(1)=0f'(x)=\dfrac{1-\ln x}{x^{2}},\ x>0,\ f(1)=0 | (ii) x=2x=2 | (Q) 1e\dfrac1e |
| (III) f′(x)=x(2−x)e−x, x>0, lim⁡x→0+f(x)=0f'(x)=x(2-x)e^{-x},\ x>0,\ \lim_{x\to0^{+}}f(x)=0 | (iii) x=1ex=\dfrac1e | (R) ee |
| (IV) f′(x)=(x−1)exx2, x>0, f(1)=ef'(x)=\dfrac{(x-1)e^{x}}{x^{2}},\ x>0,\ f(1)=e | (iv) x=ex=e | (S) −1e-\dfrac1e |

Question 1

Single correct option

Which function of Column 1 has a negative global extreme value? The correct combination is
  1. A. (I) (iii) (S)
  2. B. (I) (iv) (S)
  3. C. (I) (iii) (Q)
  4. D. (IV) (iii) (S)
Show answer and solution

Answer: (A) (I) (iii) (S)

Integrate the derivative with its initial value: f′(x)=1+ln⁡xf'(x)=1+\ln x with f(1)=0f(1)=0 gives f(x)=xln⁡xf(x)=x\ln x. Then f′(x)=0f'(x)=0 at x=1ex=\frac1e, and f′′(x)=1x>0f''(x)=\frac1x>0, so this is the global minimum, with f(1e)=−1ef\left(\frac1e\right)=-\frac1e. This is the only function in the table whose extreme value is negative: the others give 1e\frac1e (for ln⁡x/x\ln x/x), 4e−24e^{-2} (for x2e−xx^2e^{-x}) and ee (for ex/xe^x/x), all positive. Matching entries, the abscissa 1e\frac1e is (iii) and the value −1e-\frac1e is (S). So (I) →\to (iii) →\to (S). The other options either use the abscissa of a different function or the extreme value of another row.

Question 2

Single correct option

Consider the function of Column 1 which tends to 00 as x→∞x\to\infty and whose global maximum occurs at a point with x>2x>2. The correct combination is
  1. A. (III) (ii) (Q)
  2. B. (II) (iv) (P)
  3. C. (II) (iv) (Q)
  4. D. (II) (i) (Q)
Show answer and solution

Answer: (C) (II) (iv) (Q)

Integrate each derivative: ∫1−ln⁡xx2dx=ln⁡xx+C\int\frac{1-\ln x}{x^2}dx=\frac{\ln x}{x}+C with f(1)=0f(1)=0 gives f=ln⁡xxf=\frac{\ln x}{x}; and ∫x(2−x)e−xdx=x2e−x+C\int x(2-x)e^{-x}dx=x^2e^{-x}+C with f(x)→0f(x)\to0 as x→0+x\to0^+ gives f=x2e−xf=x^2e^{-x}. Both tend to 00 as x→∞x\to\infty, while xln⁡xx\ln x and exx\frac{e^x}{x} tend to ∞\infty. The maximum of ln⁡xx\frac{\ln x}{x} is at x=e≈2.718>2x=e\approx2.718>2 with value 1e\frac1e; the maximum of x2e−xx^2e^{-x} is at exactly x=2x=2, not greater than 22, with value 4e−24e^{-2}. So the function is (II), with abscissa (iv) x=ex=e and value (Q) 1e\frac1e. Hence (II) →\to (iv) →\to (Q). Thus the answer is the combination built on the logarithmic function with extremum at x=ex=e.

Question 3

Single correct option

The function of Column 1 whose global extreme value equals 4e−24e^{-2} is paired as follows. The correct combination is
  1. A. (III) (i) (P)
  2. B. (III) (ii) (P)
  3. C. (IV) (ii) (P)
  4. D. (III) (ii) (R)
Show answer and solution

Answer: (B) (III) (ii) (P)

Take f′(x)=x(2−x)e−xf'(x)=x(2-x)e^{-x} with f(x)→0f(x)\to0 as x→0+x\to0^+. Since ddx(x2e−x)=(2x−x2)e−x\frac{d}{dx}\left(x^2e^{-x}\right)=(2x-x^2)e^{-x}, we get f(x)=x2e−xf(x)=x^2e^{-x}. Then f′(x)>0f'(x)>0 on (0,2)(0,2) and f′(x)<0f'(x)<0 for x>2x>2, so x=2x=2 is the global maximum on (0,∞)(0,\infty) (the infimum 00 is not attained, so this is the only global extremum), and f(2)=4e−2≈0.54f(2)=4e^{-2}\approx0.54. The row for (IV) has f=exxf=\frac{e^x}{x}, with global minimum ee at x=1x=1, so it cannot be the maximum 4e−24e^{-2}. Checking the table: abscissa x=2x=2 is (ii), value 4e−24e^{-2} is (P). Therefore (III) →\to (ii) →\to (P). The wrong options either place the abscissa x=1x=1, belonging to the function (IV), or the value ee with this function. Thus the answer is the combination built on the function x2e−xx^2e^{-x}.
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